Question Video: Finding the First Derivative of a Composition of Radical and Trigonometric Functions Using the Quotient Rule Mathematics • Higher Education

Video Transcript

Determine the derivative of the function 𝑠 of 𝑡 equals the square root of negative sin 𝑡 plus seven over negative cos 𝑡 plus seven.

Now, we can see straightaway in this question that we have a quotient. So, we’re going to need to apply the quotient rule at some point. But are we going to need to do anything else? Well, we don’t just have this quotient. We have the square root of this quotient, which means we have a composite function. And so, we’re also going to need to apply the chain rule. We’ll begin by allowing 𝑢 to be that quotient underneath the square root. 𝑢 is equal to negative sin 𝑡 plus seven over negative cos 𝑡 plus seven. Then, 𝑠 becomes a function of 𝑢. It’s equal to the square root of 𝑢, which we can express using index notation as 𝑢 to the power of one-half.

The chain rule, using the letters 𝑠, 𝑡, and 𝑢 as we have in this question, tells us that the derivative of 𝑠 with the respect to 𝑡 is equal to d𝑠 by d𝑢 multiplied by d𝑢 by d𝑡. Applying the power rule, we see that d𝑠 by d𝑢 is equal to one-half 𝑢 to the power of negative a half. But in order to find d𝑢 by d𝑡, we’re going to need to apply the quotient rule.

We’ll define 𝑓 to be the function in the numerator, that’s negative sin 𝑡 plus seven, and 𝑔 to be the function in the denominator, negative cos 𝑡 plus seven. In order to find the derivatives of these two functions, we need to recall how we differentiate sin and cos. There is a helpful little cycle that we can remember. The derivative of sin 𝑡 is cos 𝑡. The derivative of cos 𝑡 is negative sin 𝑡. The derivative of negative sin 𝑡 is negative cos 𝑡. And the derivative of negative cos 𝑡 is sin 𝑡. And then we go around the cycle again.

Remembering that the derivative of a constant is just zero, we have that 𝑓 prime is equal to negative cos 𝑡 and 𝑔 prime is equal to sin 𝑡. Now, we can substitute into the quotient rule to find d𝑢 by d𝑡. It’s 𝑔, that’s negative cos 𝑡 plus seven, multiplied by 𝑓 prime, negative cos 𝑡, minus 𝑓, that’s negative sin 𝑡 plus seven, multiplied by 𝑔 prime, that’s sin 𝑡, all over 𝑔 squared. Now, we need to do some simplification. So, we’ll expand the parentheses in the numerator, which gives cos squared 𝑡 minus seven cos 𝑡 plus sin squared 𝑡 minus seven sin 𝑡 over negative cos 𝑡 plus seven squared.

We can recall, at this point, one of our trigonometric identities, cos squared 𝑡 plus sin squared 𝑡 is equal to one. So, this simplifies to one minus seven cos 𝑡 minus seven sin 𝑡 over negative cos 𝑡 plus seven squared. Now that we found both d𝑢 by d𝑡 and d𝑠 by d𝑢, we can substitute into the chain rule. We have then that d𝑠 by d𝑡 is equal to d𝑠 by d𝑢, that’s a half 𝑢 to the power of negative a half, multiplied by d𝑢 by d𝑡, that’s one minus seven cos 𝑡 minus seven sin 𝑡 over negative cos 𝑡 plus seven squared.

Remember, though, that d𝑠 by d𝑡 must be in terms of 𝑡 only. So, we need to reverse our substitution. We have negative sin 𝑡 plus seven over negative cos 𝑡 plus seven to the power of negative a half multiplied by one minus seven cos 𝑡 minus seven sin 𝑡 over two multiplied by negative cos 𝑡 plus seven squared. That power of negative a half means a reciprocal, so we can deal with that by inverting the fraction. And the first part becomes negative cos 𝑡 plus seven over negative sin 𝑡 plus seven to the power of positive one-half.

We can then simplify the powers. We have negative cos 𝑡 plus seven to the power of a half in the numerator and then negative cos 𝑡 plus seven to the power of two in the denominator. Which will lead to a power of negative three over two overall. That’s a power of three over two in the denominator. This leads us to one minus seven cos 𝑡 minus seven sin 𝑡 in the numerator. And in the denominator two times the square root of negative sin 𝑡 plus seven. That’s negative sin 𝑡 plus seven to the power of a half multiplied by negative cos 𝑡 plus seven to the power of three over two.

In this question then, we’ve seen that we can apply both the quotient and chain rule to a problem involving the derivatives of trigonometric functions.